ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 0 August 2005 Final Examination R. Culham & M. Bahrami This is a 2 - /2 hour, closed-book examination. You are permitted to use one 8.5 in. in. crib sheet (both sides), Conversion Factors (inside cover of text) and the Property Tables and Figures from your text book. There are 5 questions to be answered. Read the questions very carefully. Clearly state all assumptions. It is your responsibility to write clearly and legibly. hen using correlations, it is your responsibility to verify that all limiting conditions are satisfied. Question (2 marks) Steam flows through a stainless steel pipe and maintains the temperature of the inside surface of the pipe at T 00 C. The pipe is enclosed in a layer of rubber as shown in the sketch. Because of the imperfect contact between the rubber and the stainless steel, a thermal contact resistance must be considered at this interface. The dimensions are r 25mm, r 2 35mm and r 3 55mm. The assembly is exposed to the environment at T and the corresponding surface heat transfer coefficient is h 2/(m 2 K). The conductivity of the stainless steel is k ss 5/(m K) and the conductivity of the rubber is k rub 0.5 /(m K). Two thermocouples provide two temperature measurements - one at the surface (at r r 3 ) where T 3 40 C and the other embedded half way through the rubber (at r (r 2 + r 3 )/2) where T m 65 C. Determine: a) the rate of heat loss per unit length of the pipe, [/m] b) the temperature of the environment, [ C] c) the specific thermal contact resistance between the pipe and the rubber, [K m 2 / ]
Part a) Since we need to find the midpoint temperature in the rubber, break the rubber into two concentric rings each with an equivalent radius. R total R ss + R c + R rub + R rub2 + R conv R ss ln(r 2 r ) 2πLk ss R rub ln(r m r 2 ) 2πLk rub R rub2 ln(r 3 r m ) 2πLk rub ln(0.035/0.025) 2πL(m) 5 (/m K) 0.00357 ( ) K L ln(0.045/0.035) 2πL(m) 0.5 (/m K) 0.2667 ( ) K L ln(0.055/0.045) 2πL(m) 0.5 (/m K) 0.229 ( ) K L R conv ha 2πr 3 Lh 2π 0.055(m) L 2 (/m 2 K) 0.24 ( ) K L Since the heat flow is the same throughout each section of the pipe, the heat loss per unit length can be written as Q T m T 3 R rub2 L (65 40) K ( ) 0.229 K L(m) L 7.426 m Part b) The environmental temperature can be determined from solving for T Q T R conv (T 3 T ) /(2π r 3 h) (2π r 3 h) (T 3 T ) Q 7.426 /m T T 3 2π r 3 h 40 C 2π 0.055 (m) 2 (/m 2 K).69 C
Part c) Since we know temperatures at T and T m we need to find the resistance between these two points. Then the contact resistance can be written as R c R m R ss R rub First, Q T T m R m L R m T T m Q L (00 65) K 7.426 (/m) L (m) 0.298 L ( ) K Therefore R c 0.298 L ( ) K 0.00357 L ( ) K 0.2667 L ( ) K 0.02783 L ( ) K and R c R c A R c (2π r 2 L) 0.02783 ( ) K 2π 0.035 (m) L (m) L 0.006 K m2
Question 2 (2 marks) a) As the length of a fin increases, would you expect i) the fin efficiency and ii) the fin effectiveness to increase or decrease? Explain. (3 marks) b) Oil is driven over a heated surface where hydrodynamic and thermal boundary layers are formed. The flow is laminar. hat fluid property can be used as an indication of the relative thickness of the two boundary layers? Explain why. hich boundary layer is thicker? hy? (3 marks) c) If 0% of the emitted radiant energy from an incandescent light bulb falls in the visible wavelength band (i.e. between 0.4 and 0.76 µm), determine the operating temperature of the filament. (4 marks) d) How would the value of the critical radius of insulation change if radiation effects are considered? Explain. (2 marks) Part a) i) The finefficiency can be written as η tanh ml For a fixed value of the fin parameter, m, you will note that (tanh L)/L will always decrease as L is increased. This means that the finefficiency will decrease for an increase in the fin length. or ( From Fig. 8-59, we see that ξ L + ) 2 t h/kt. AsL, theefficiency, η. ml ii) The fin effectiveness is given as ɛ Q fin Q no f in As the fin length, L, increases, the overall heat transfer of the fin, Q fin will also increase. Therefore, the fin effectiveness will increase foranincreaseinthefin length. Part b) The Prandtl number is the property that gives a clear indication of the relative thickness of the boundary layers. Pr ν α rate of momentum diffusion rate of thermal diffusion The Prandtl number for oil is between 84-4700 depending on the temperature. (Table A-8). This means that momentum is diffusing faster than heat and the hydrodynamic boundary layer will be thicker than the thermal boundary layer.
Part c) From the general formulation for the blackbody radiation function we can write λ2 E bλ (T ) dλ λ f λ (T ) f λ2 (T ) σt 4 0. Since T is unknown, we will have to iterate to find a value of T that satisfies the above equation. From Table 2.2: at T 3000 K λ T 0.4 µm 3000 K 200 µm K f 0 0.4 0.00234 λ 2 T 0.76 µm 3000 K 2280 µm K f 0 0.7 0.6635 f 0.4 0.76 (3000 K) 0.6635 0.00234 0.450 slightly high at T 2900 K λ T 0.4 µm 2900 K 60 µm K f 0 0.4 0.0077 λ 2 T 0.76 µm 2900 K 2204 µm K f 0 0.7 0.0675 f 0.4 0.76 (2900 K) 0.0675 0.0077 0.099904 9.99% 0% The filament temperature would be 2900 K Part d) The critical radius of insulation can be written as: r crit k h cylinder and r crit 2k h sphere If radiation effects are included this would result in a higher effective heat transfer coefficient such that h eff h cond + h rad where h rad ɛσ(t s + T sur )(T 2 s + T 2 sur ) A higher effective heat transfer coefficient will always lead to a lower value of critical radius of insulation.
Question 3 (3 marks) A radiant space heater consists of a long, diffuse, gray, cylindrical heating element that has a diameter of 30 mm and is backed by a curved metallic reflector. The surface temperature and emissivity of the heating element are T 945 K and ɛ 0.8. The temperature of the air and the temperature of the surroundings are both 300 K. Convection heat transfer from the element can be considered to be negligible. The reflector, which has an arc length of S 0.3 m, may be assumed to be an isothermal, diffuse, gray surface with an emissivity of ɛ 2 0.. The base of the reflector has a width of 0.5 m. The steady state temperature of the reflector plate is T 2 385 K and the view factor of the heater with respect to the reflector is F 2 0.625. a) Draw the equivalent circuit to describe radiation heat transfer between the reflector, the heating element and the surroundings. Omit the portion of the circuit describing heat transfer from the back of the reflector. b) Evaluate all resistances and nodal potentials (per unit length of the heater). c) Calculate the electrical power input to the heating element per unit length.
Assume:. surface are gray and diffuse 2. the surroundings form a large enclosure which may be represented by a hypothetical surface of temperature T 3 T sur and emissivity, ɛ 3 Part a)
Part b) The view factors are obtained as follows: F 3 F 2 0.625 0.375 (summation rule) ( ) ( ) A πdl F 32 F 3 F 3 0.375 L The circuit resistances are ɛ ɛ A A 3 0.8 0.8 π 0.03 (m) L (m) 2.65 L (m 2 ) 0.375 π 0.03 m 0.5 m 0.764 A F 2 A F 3 A 2 F 23 ɛ 2 ɛ 2 A 2 π 0.03 (m) L (m) 0.625 6.98 L (m 2 ) π 0.03 (m) L (m) 0.375 28.30 L (m 2 ) A 3 F 32 0.5 (m) L (m) 0.764 8.73 L (m 2 ) 0. 0. 0.30 (m) L (m) 30 L (m 2 ) The nodal potentials are ( ) E b σt 4 5.67 0 8 (945 K) 4 45, 28 (/m 2 ) m 2 K 4 ( ) E b2 σt 4 2 5.67 0 8 (385 K) 4, 246 (/m 2 ) m 2 K 4 ( ) E b3 σt 4 3 5.67 0 8 (300 K) 4 459 (/m 2 ) m 2 K 4 Part c) The input power can be obtained as follows Q E b J ɛ ɛ A First we need to find the radiosity by applying radiation balances to J and J 2. E b J ɛ J J 2 + J E b3 ɛ A 45, 28 J 2.65/L A F 2 A F 3 J J 2 6.98/L + J 459 28.30/L J 2 8.0J 290, 0.93
and E b2 J 2 ɛ 2 ɛ 2 A 2 J 2 J A F 2 + J 2 E b3 A 2 F 23, 246 J 2 30/L J 2 J 6.98/L + J 2 459 8.73/L J 3.5J 2 597.99 J 3.5(8.0J 290, 0.93) 597.99 J 37, 600 /m 2 hich in turn gives J 2 8.0J 290, 0.93 8.0(37, 600) 290, 0.93, 64 /m 2 and finally power input per unit length is Q L E b J ɛ ɛ A L (45, 28 37, 600) /m2 2.65 L (m 2 ) L (m) 2, 874.7 (/m)
Question 4 (0 marks) A steel bar, initially at an ambient temperature of 25 C, is heat treated by suspending it vertically in the lengthwise direction and passing it through a conveyor oven maintained at a temperature of 75 C. The steel bar has dimensions of 32 mm 0 mm. m and thermophysical properties as follows: ρ 83 kg/m 3 C p 434 J/(kg K) k 4/(m K) α.6 0 6 m 2 /s In order to achieve the proper heat treatment, the steel bar must reach a temperature of at least 40 C for 35 min. If all thermophysical properties and oven conditions remain the same, how long should a 76 mm 35 mm.6 m steel bar remain in the oven to achieve the proper heat treatment at 40 C? Since we do not have a value for the heat transfer coefficient, h, we will have to assume that Bi < 0. and then check later to see if the assumption is correct. First examine the smaller bar to determine the value of h. The characteristic length can be determined as L V A 0.032 m 0.00 m. 2(0.032 m 0.00 m) + 2(0.032 m. m) + 2(0.00 m. m) 0.00378 m Using a lumped capacitance procedure, we can write where T (t) T T i T e t/τ τ mc p Ah ρv C p ha ρlc p h 83 kg/m3 0.00378 m 434 J/(kg K) h (/m 2 K) Therefore 3, 339. (s) h (40 75) C (25 75) C exp min 60 s/min 35 3, 339. (s) h 0.2333 exp( 0.574h)
and h 9.25 /m 2 K Check the Biot number to see if a lumped system approach was correct. Bi hl k 9.25 /m2 K 0.00378 m 4 /(m K) 0.0009 Since Bi << 0. the lumped system approach is valid. The characteristic length for the larger steel bar is L V A 0.076 m 0.035 m.6 2(0.076 m 0.035 m) + 2(0.076 m.6 m) + 2(0.035 m.6 m) 0.08 m Therefore τ ρlc p h 83 kg/m3 0.08 m 434 J/(kg K) 9.25 /m 2 K (40 75) ( C (25 75) C exp t(s) ) 450.7 (s) 0.2333 exp( t/450.7) 450.7 s t 655.9 s 09.2 minutes Check again to see if the Biot number is small. Bi hl k 9.25 /m2 K 0.08 m 4 /m K 0.00266 << 0.
Question 5 (3 marks) Consider two 0.5 m 0.7 m 0.05 m horizontal, parallel plates separated by a distance s in a room at a constant temperature of T 30 C. The plates are made of iron (Armco, 99% pure). The top surface of the lower plate is maintained at a temperature of 74 C while the bottom surface of the upper plate is maintained at 4 C. a) The distance between the plates, s, is large so that the space between the plates, can be assumed to be a constant temperature at T 30 C. Determine the net convection heat transfer rate from the inner surfaces of the plates. b) The distance between the plates is s 0. m. Assume -D heat flow between the plates (through the air gap). Determine an equivalent effective thermal conductivity for the overall system (where the system includes the plates and the air gap. You can use the same convective heat transfer coefficients, h and h 2 from part a). Neglect radiation effects. Assume: steady state heat transfer air pressure is at atm -D conduction The properties of the air are evaluated at the film temperature: T f T + T 2 T f2 T 2 + T 2 Air properties from Table A-9 74 + 30 2 4+30 2 52 C 325 K 7 C 290 K Surface Surface 2 k 0.0279 /m K k 2 0.0253 /m K Pr 0.709 Pr 2 0.74 ν.85 0 5 m 2 /s ν 2.48 0 5 m 2 /s β T f 325 K 0.00308 K β 2 0.00345 K T f 290 K
For the horizontal, hot surface facing up, the characteristic length and the Rayleigh number are: δ A P (0.5 m)(0.7 m) 2(0.5 +0.7) m 0.458 m From Eq. - Ra gβ (T T )δ 3 Pr ν 2 ( 9.8 m ) ( 0.00308 ) (74 30) K (0.458 m) 3 s 2 K (0.709) (.85 0 5 m 2 /s) 2 8.86 0 6 Nu 0.54Ra /4 0.54(8.86 0 6 ) /4 29.46 Nu h δ 29.45 0.0279 /m K 29.46 h 5.64 /m 2 K k 0.458 m For the lower surface of a horizontal cold plate, the characteristic length and the Rayleigh number are: From Eq. - δ A P (0.5 m)(0.7 m) 2(0.5 +0.7) m 0.458 m Ra 2 gβ 2(T T 2 )δ 3 Pr 2 ν 2 2 ( 9.8 m ) ( 0.00345 ) (30 4) K (0.458 m) 3 s 2 K (0.74) (.48 0 5 m 2 /s) 2 8.88 0 6 Nu 2 0.54Ra /4 2 0.54(8.88 0 6 ) /4 29.48 Nu 2 h 2δ 29.48 h 2 k 2 The net heat transfer can be written as 29.48 0.0253 /m K 0.458 m 5.2 /m 2 K Q net Q Q 2 A[h (T T ) h 2 (T T 2 )] (0.5 m 0.7 m)[(5.64 /m 2 K)(74 30) K (5.2 /m 2 K)(30 4) K] 40.264
Part b) k iron 72.7 /m K Table A-4 k air (@32 K) 0.02694 /m K Table A-9 The total resistance in the system can be written as a series resistance path where R system R plate + R conv + R air + R conv2 + R plate2 t k A + h A + s k air A + h 2 A + t 2 k 2 A 0.05 m 0. m + + A 72.7 5.64 0.02694 m K m 2 K m K + + 0.05 m 5.2 72.7 m 2 K m K 4.0853/A The effective resistance can be written as s +2t k eff A 4.0853 A k eff 0. m + 2(0.05 m) 0.0490 /m K 4.0853